On the idea of time in physics-relativity

Discussion in 'Physics & Math' started by ash64449, Mar 15, 2013.

  1. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    The internet observer must remember that the diagram shows the platform reference frame. It gets its coordinates from platform clocks and platform rulers, so it only shows the speed of the flashes relative to the platform.

    If they want to cdetermine the speed relative to the train, they need to use the train clocks and train rulers, which are shown in the other diagram:

    Please Register or Log in to view the hidden image!


    Relative to the train:
    The yellow flash begins at t'=0, x'=0, and reaches the front of the train at t' = 12.5, x' = 12.5.
    It travels 12.5 units distance in 12.5 units time. Its speed is 1.
    The red flash begins at t'=0, x'=0, and reaches the back of the train at t' = 12.5, x' = -12.5.
    It travels -12.5 units distance in 12.5 units time. Its speed is -1.

    EDIT: These numbers are wrong. I looked at the blue flashes instead of the red and yellow. Correct numbers are below.
     
    Last edited: Apr 5, 2013
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  3. Neddy Bate Valued Senior Member

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    I think you meant to say the blue flashes, not the yellow and red flashes.
     
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  5. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    Whoops... I meant red and yellow flashes, but looked at blue flashes. Thanks for that.

    Relative to the train:
    The yellow flash begins at t'=7.5, x'=-12.5, and reaches the front of the train at t' = 32.5, x' = 12.5.
    It travels 25 units distance in 25 units time. Its speed is 1.
    The red flash begins at t'=-7.5, x'=12.5, and reaches the back of the train at t' = 17.5, x' = -12.5.
    It travels -25 units distance in 25 units time. Its speed is -1.
     
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  7. Prof.Layman totally internally reflected Registered Senior Member

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    982
    IDK, I think we have probably nit picked at it enough for now. I have been thinking about it and it would seem that the train should be able to gain more or less on the beams from the platforms frame so that it measures the same speed of light. But, then the observer on the train would not see the beams gain more or less on the train. So then how would the platform see the light gain on the train and the train not see the light gain on the train? (one beam gaining on the train more or less than the other) It would seem like they both wouldn't be able to agree if the beam is gaining on the train or not. The beam couldn't gain on the train and not gain on the train at the same time. But, then you would think that it has to inorder to be the speed of light in both frames. Then I think I just hit another stumbling block that makes it where I think it just cannot be described with these simple diagrams.
     
  8. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    The way past the stumbling block is to see what Einstein did - that simultaneity is relative.
    Clocks that are synchronized in the platform frame are not synchronized in the train frame, and vice versa, which means that the two reference frames disagree about whether two things in different places happen at the same time.

    You like Sean Carroll, right? Read this: Time-saving tips for understanding Einstein
    Here’s the truth: Einstein proposed that the amount of time elapsed between two spacetime events depends (in a very definite way) on the path taken between those events. It is not simply a universal constant, as it would be in Newtonian physics. So the notion of “simultaneity” for distant events is just a useful approximation, valid when everyone is traveling slowly compared to the speed of light. And you know what? Einstein was right. It’s been verified over and over again, from the lifetimes of rapidly-moving subatomic particles to the time kept by atomic clocks moving in airplanes. Deal with it.

    You might also be interested in his book From Eternity to Here, which seems to cover the relativity of simultaneity in Chapter 4:
    Starting from a single event in Newtonian spacetime, we were able to define a surface of constant time that spread uniquely throughout the universe, splitting the set of all events into the past and the future (plus “simultaneous” events precisely on the surface). In relativity we can’t do that.
     
  9. ash64449 Registered Senior Member

    Messages:
    795
    PL, I will try to make you understand that MME works on the basis Of TE. But it makes use of ether and all,to show history.But at last it is explained why it is meaningless to use ether. i will provide lengthy explanation now. Then you will understand. It has some series of videos to make you understand.
     
  10. ash64449 Registered Senior Member

    Messages:
    795
    Here is the video and series of explanations:

    PLEASE NOTE: When replaying these animations, click on the circular arrow in the bottom left corner of the video window. If you click inside the video, you may end up running a different video.

    We start during the era before Einstein when most of the great scientists believed in an absolute ether rest state in which light propagated at the same constant speed in all directions. Imagine a very brief bright flash of light being set off in this stationary ether. It will create an ever-expanding spherical shell of light, centered on its point of origin with respect to the stationary ether. Here we picture a stick-man setting off a flash of blue light creating a wave much like you would see dropping a pepple in a pool of water.

    [YOUTUBE]U0SC5lY2LZ0[/YOUTUBE]

    They believed that if the man were moving with respect to this stationary ether frame, he would not remain in the center of this expanding spherical shell but would move off-center.

    [YOUTUBE]TtLOm7B6YMA[/YOUTUBE]

    But the question is: how can the man tell if he remains in the center of this expanding shell or moves off-center? By analogy, we could visualize what would happen if we were observing an expanding ring of waves on the surface of a pool after dropping a pebble in the water because we use light to observe the water, but how can we observe a lightwave once it has started moving away from us? Therein lies the problem: we cannot directly observe the propagation of light so we do the next best thing which is to set up an array of mirrors to reflect the light back to us.

    Now the best way for the man to "observe" an expanding spherical shell of light is to set up a whole bunch of mirrors, all an equal distance from him and in all possible directions. Then when he sets off the flash it will expand until it simultaneously hits all the mirrors which turn the expanding spherical shell of light into a contracting spherical shell of light which will eventually collapse on the man simultaneously from all directions.

    For purposes of illustration, we will consider a two-dimensional subset of mirrors and an expanding ring of light, much like the expanding ring of waves set off by a pepple dropped in the center of a circular pool of water as it simultaneously strikes the entire pool wall circumference, reverses direction and simultaneously collapses on the center of the pool.

    [YOUTUBE]jLNxE5g0pyo[/YOUTUBE]

    Now it's really not practical to build a solid ring of mirrors but all we really need is four mirrors that are placed 90 degrees apart and all equidistant from the observer. In this animation, I have used four circular mirrors so that when the light strikes them, they each create a new expanding circle of light. Notice how the four reflections arrive simultaneously on the observer.

    [YOUTUBE]ygvY4AjwPmE[/YOUTUBE]

    I represent the stationary observer in green and I call him Homer (think green, green grass of home). I represent the original expanding circle of light in blue as well as a blue dot to represent its source, the mirrors in yellow, the collapsing circles of light in green when they reflect off stationary mirrors.

    Please note that just as in the previous post when the collapsing circle of light arrived simultaneously from all directions on the observer, the four reflections from the four mirrors all arrive simultaneously on the observer. Although this is not actually how the Michelson-Morley Experiment (MMX) was configured, it still represents conceptually exactly what the experiment was doing.

    The MMX experimenters assumed that the previous animation would represent only what would happen if they were stationary with respect to the ether which they assumed they never were. They believed that they were constantly moving with respect to the ether and also constantly changing their velocity through the ether as the earth rotated on its axis and as it revolved around the sun. This constant acceleration was very small so for all practical purposes, they could assume that they were moving at a constant speed through the ether during the brief time interval of the experiment. This is how they thought the light would behave and note that now all four reflection do not reach the man at the same time:

    [YOUTUBE]U625Pjm9M-I[/YOUTUBE]

    I represent the moving observer in red and I call him Rover (think Red Rover). The light that reflects off the moving mirrors is shown in red and a red dot is placed at the origin of each expanding reflection.

    Note that when the light from the four mirrors arrives at Rover, it is not simultaneous, it first arrives from the top and bottom mirrors and then later arrives from the left and right mirrors. This is what the MMX experimenters expected to measure but instead, they got the same result as if they were stationary in the ether, the same result that Homer would have gotten.

    So now the question is how can this happen? Well, Lorentz and others came up with an explanation and we will go through a process that will arrive at the same explanation.

    First, we want to learn how we know where to put the mirrors so that the expanding circle of light can create a reflection that results in a collapsing circle of light in just the right place at just the right time. For Homer, it's easy:

    [YOUTUBE]Y0XWb6Il92A[/YOUTUBE]

    Just note the intersection of the blue expanding circle and the green collapsing circle and in this animation, we draw a black dashed line to show where that intersection occurs:

    [YOUTUBE]7VftpL1KL6o[/YOUTUBE]

    Now for Rover, it's a little more complicated because his collapsing circle of light is not centered on the expanding circle of light but rather the location of where he will be later on, shown as a red dot. Try to visualize in this animation where the blue and red circles intersect:

    [YOUTUBE]dHO6jtkDXM8[/YOUTUBE]

    And here we have the black dashed line to show the points of reflection:

    [YOUTUBE]q9qJhCaHotI[/YOUTUBE]

    Now this black dashed line shows the points of relection relative to the ether but we really want them relative to Rover, so here we show both for comparison:

    [YOUTUBE]sE0G1wcvrRI[/YOUTUBE]

    Also, note that Lorentz realizes that everything contracts in the direction of motion so we now show Rover as being length contracted as well as his arrangement of mirrors. In addition, the time it takes for the light to traverse from Rover to the mirrors and back to Rover is longer than it was for Homer which illustrates time dilation. We can also see the issue of Relativity of Simultaneity because the reflections for Rover do not all occur at the same time whereas they do for Homer.

    [YOUTUBE]S7r5GeIfZas[/YOUTUBE]

    This illustrates how Lorentz believed MMX produced the null result. He believed that the experiment was moving through the ether and experienced length contraction, time dilation and relativity of simultaneity.

    He also believed that Rover would measure the speed of light to be the same as Homer because even though time was going slower and stretching out (time dilation), it is the actual length that the light has to travel relative to the ether that is used to calculate the speed (length divided by time), so we need to use the lengths defined by the black dashed line, not the moving brown line representing the length contracted mirror. This length is dilated to the same extent that time is dilated and so the two dilations cancel each other out and give the same calculation for the speed of light.

    However, Einstein put a new spin on the interpretation. He said that we could assume that MMX was actually stationary in the ether and everything else that was moving with respect to MMX was experiencing length contraction, time dilation and relativity of simultaneity.

    Now I want to show what would happen if instead of a solid ring of mirrors, we used individual mirrors with gaps between them. This will allow some of the light to be reflected and some to pass through. First we go back to Homer:

    [YOUTUBE]VU_812f0JtI[/YOUTUBE]

    Now we do the same thing for Rover:

    [YOUTUBE]_edUM0wBLsY[/YOUTUBE]

    And finally, we show combine the two of them:

    [YOUTUBE]dEhvU31YaCw[/YOUTUBE]

    This shows how two observers moving with respect to each other can both conclude that they are each in the center of the same expanding sphere of light that was emitted when they were colocated.

    An important concept to learn through this series of animations is that even though we see that Homer is stationary in the ether and Rover is moving, they themselves cannot tell the difference. Whatever Rover does to measure the positions of his mirrors, he will conclude that they are in a perfect circle since his ruler will contract when he measures the shortened distances. And whatever clock he carries will be time dilated to the same extent as the time it takes for the reflection to return to him. And of course, he has no way of knowing that the light is not reflecting off his mirrors simultaneously.

    When Einstein came along, he re-interpreted the ideas of Lorentz by saying that Rover could consider himself to be "stationary in the ether" so that his ruler was not length contracted, and his clock was not time dilated, and the light reflects off his mirrors simultaneously, and all those things happen to Homer instead. Thus he showed that the concept of an absolute ether rest state was useless as any inertial state is indistinguishable from it.

    For those that are familiar with Special Relativity, view these animations from the stand point of a Frame of Reference.
     
  11. ash64449 Registered Senior Member

    Messages:
    795
    If you cannot see the video,then go to youtube and paste the url and watch it there..
     
  12. ash64449 Registered Senior Member

    Messages:
    795
    You didn't understand what i actually said. I said that you were mixing the point of view seen in train with point of view seen in platform and you say that you see platform as large from train's point of view. You got platform as large by mixing them.. So you cannot say that i have missed a point in relativity. You didn't too missed the point but you just mixed without knowing that you mixed it..
     
  13. Prof.Layman totally internally reflected Registered Senior Member

    Messages:
    982
    Yes, but the beams of light in the MME arrive at the same time in both frames. It has been published in over a dozen or so popular science books, by writers with Ph.D.s'. In some of these books it even says that Einstein didn't know about the MME and its results, before he published that book on simultaneity, and this happened even though he published that book after the results to the MME came out. All of science was wrong when the results of the MME came out, and they discovered that there was no aether like they expected. It was a failed experiment. It even states this in the wiki. But, he was right about finding the equation for the proper time in his 1905 paper.

    If you wanted to measure the velocity of a beam of light you would say that it travels a distance that is the speed of light times time.

    \( d = c t \)

    This would be the distance a beam of light was observed to travel. If you wanted to figure out how far a beam of light has traveled in another frame, then that distance would also be the speed of light times time. But, the distance and time would have a prime symbol because it is something other than the distance in the other frame. The speed of light wouldn't be prime because it is the same in both frames.

    \( d' = c t' \)

    Then having different prime symbols on distance and time, both frames can then measure the same velocity for the speed of light of that beam. So then to find the realtion between these two frames in order to figure out the difference's of both frames time and distance, you could use the Pythagorean Theorem.

    \( ( v t )^{2} + ( c t' )^{2} = ( c t )^{2} \)

    \( c^{2}t'^{2} = (c^{2}t^{2} - v^{2}t^{2}) \)

    \( c^{2}t'^{2} = c^{2}t^{2} ( 1 - \frac{v^{2}}{c^{2}}) \)

    \( t'^{2} = t^{2}( 1 - \frac{v^{2}}{c^{2}})\)

    \( t' = t sqrt{ 1 - \frac{v^{2}}{c^{2}} \)

    So then the proper time is the difference in time of the speed of light measurments of two observers in two different frames at the same time. Their times are not equal to each other when they both measure a beam of light to be at the same location at the "same time" or when those events would be simultaneous, that would be when those two clocks measure different times. That is what happens when two observers measure the same beam of light at the "same time".
     
  14. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    I agree.
    But the train thought experiment beams are not the same as the MME beams.
    The red and yellow flashes are not the blue flashes.

    The train thought experiment does not say that the MME beams arrive at different times.
     
  15. Prof.Layman totally internally reflected Registered Senior Member

    Messages:
    982
    I don't see how you can get to the Lorentz Transformation that you used though. Say for instance you take the Galilean Transformation \( x' = x - vt \) , and then you solve for velocity.

    \( x' = x - vt \)

    \( x' + vt = x \)

    \( vt = x - x' \)

    \( v = \frac {x-x'}{t}\)

    Then you replace this with your new velocity of the ship in the Pythagorean Theorem of your beam of light measurments.

    \( c^{2} t'^{2} + \frac {(x-x')^{2}t^{2}}{t^{2}} = c^{2}t^{2} \)

    \( c^{2} t'^{2} = c^{2}t^{2} - (x-x')^{2} \)

    \( c^{2} t'^{2} = c^{2} ( t^{2} - \frac {(x-x')^{2}}{c^{2}}) \)

    \( t' = sqrt {t^{2} - \frac {(x-x')^{2}}{c^{2}} \)

    Wouldn't you have to do this type of operation in order to find the difference then of two locations from two different frames?
     
  16. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    I'm not sure what approach you're taking, but there are plenty of ways to derive the lorentz transform that Google will show you. I can't remember which way I first learned it.

    I'll perhaps draw up a graphical way if I find time.
     
  17. Robittybob1 Banned Banned

    Messages:
    4,199
    If you reckon I'm mixed up quote me so I know what you are talking about. If you reckon I'm mixed up before, I'm definitely mixed up now.
     
  18. Prof.Layman totally internally reflected Registered Senior Member

    Messages:
    982
    I was just trying to figure how there could be a boost in the "x" direction and then both frames would measure the same speed of light. It seems like the boost in the "x" direction would then have to go through the Pythagorean Theorem in order to make one frames light beam not gain more or less on another frames beam of light, so that they both measure the same speed of light in both frames. Then the velocity would be measured from the platforms frame, so then the observer on the platform would be using his rulers and clocks in order to determine the trains velocity. I think it would be interesting to see a graph of this "proper boost". The x' would just be the distance the platform observer measured the train to travel and not the distance measured in the other frame.
     
  19. ash64449 Registered Senior Member

    Messages:
    795
    what?
     
  20. ash64449 Registered Senior Member

    Messages:
    795
    What is with you? Can't you look at the comment i said. You will never understand until you look what i said. Clearly says that.

    And one thing you should understand: Simultaneous events in one frame may not be simultaneous in other frame and vice versa
     
  21. Robittybob1 Banned Banned

    Messages:
    4,199
    You are just about impossible to understand. I'll just ignore you from now on, unless when you criticise me you quote words I've used to emphasize your points.
     
  22. Robittybob1 Banned Banned

    Messages:
    4,199
    A single pair of simultaneous events can be simultaneous in more than one frame (I proved this earlier in this thread). A series of events will not have the same simultaneity in different frames. I think this is what you were meaning. What is the correct wording of that statement? "Simultaneous events in one frame may not be simultaneous in other frame and vice versa." What did Einstein actually say?
     
  23. eram Sciengineer Valued Senior Member

    Messages:
    1,875
    Hey ash64449, I corrected the links.

    Please Register or Log in to view the hidden image!





    PLEASE NOTE: When replaying these animations, click on the circular arrow in the bottom left corner of the video window. If you click inside the video, you may end up running a different video.

    We start during the era before Einstein when most of the great scientists believed in an absolute ether rest state in which light propagated at the same constant speed in all directions. Imagine a very brief bright flash of light being set off in this stationary ether. It will create an ever-expanding spherical shell of light, centered on its point of origin with respect to the stationary ether. Here we picture a stick-man setting off a flash of blue light creating a wave much like you would see dropping a pepple in a pool of water.

    [video=youtube;U0SC5lY2LZ0]http://www.youtube.com/watch?v=U0SC5lY2LZ0[/video]

    They believed that if the man were moving with respect to this stationary ether frame, he would not remain in the center of this expanding spherical shell but would move off-center.

    [video=youtube;TtLOm7B6YMA]http://www.youtube.com/watch?v=TtLOm7B6YMA[/video]

    But the question is: how can the man tell if he remains in the center of this expanding shell or moves off-center? By analogy, we could visualize what would happen if we were observing an expanding ring of waves on the surface of a pool after dropping a pebble in the water because we use light to observe the water, but how can we observe a lightwave once it has started moving away from us? Therein lies the problem: we cannot directly observe the propagation of light so we do the next best thing which is to set up an array of mirrors to reflect the light back to us.

    Now the best way for the man to "observe" an expanding spherical shell of light is to set up a whole bunch of mirrors, all an equal distance from him and in all possible directions. Then when he sets off the flash it will expand until it simultaneously hits all the mirrors which turn the expanding spherical shell of light into a contracting spherical shell of light which will eventually collapse on the man simultaneously from all directions.

    For purposes of illustration, we will consider a two-dimensional subset of mirrors and an expanding ring of light, much like the expanding ring of waves set off by a pepple dropped in the center of a circular pool of water as it simultaneously strikes the entire pool wall circumference, reverses direction and simultaneously collapses on the center of the pool.

    [video=youtube;jLNxE5g0pyo]http://www.youtube.com/watch?v=jLNxE5g0pyo[/video]
     

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