Nature of Time Dilation and Length Contraction

[tsmid]

It is perfectly obvious that the Lorentz transformation is both linear and not consistent with vectorial velocity addition. You are simply forcing vectorial velocity addition in between the lines of Einstein's derivation. You made that clear by saying that Einstein's motivation was to be able to apply x'=x-vt, which is clearly false.

Here is the transformation that he did use:

x'-ct'=λ(x-ct)

Observe what happens when you differentiate with respect to t.

dx'/dt-c(dt'/dt)=λ[dx/dt-c(dt/dt)]
(dx'/dt')(dt'/dt)-c(dt'/dt)=λ[(dx/dt)-c]
(dt'/dt)[(dx'/dt')-c]=λ[(dx/dt)-c]

We don't have (dt'/dt) or λ yet, but we don't need them to see that this is not vectorial velocity addition. Let dx'/dt' be the speed of light in the primed frame, and let dx/dt be the speed of light in the unprimed frame. If dx/dt=c then we have the following for dx'/dt':

(dt'/dt)[(dx'/dt')-c]=λ[c-c]
(dx'/dt')-c=0
dx'/dt'=c

It could not possibly be more obvious that the linear transformation at which Einstein arrived does not imply vectorial velocity addition.

As I said before, the vectorial velocity addition only apparently disappears due to Einstein's re-scaling of the space and time coordinates. It is very much like claiming that you stay on the same spot when making a number of steps to which you have assigned the length 0 by definition.

The motivation to treat the propagation of light in the same way as that of ordinary objects (i.e. involving a frame dependent velocity) is clear from the similarity of the Galilei and Lorentz transformation

x'=x-vt and
x'=gamma*(x-vt)

Since the first equation is the limit of the second for small v (gamma=1 then) , one can hardly argue that there is a vectorial velocity addition implied with one but not the other.

You seem to be confused about the "usual definition of speed", because you say that it "implies a linear transformation in the above" (that is, Galilean) "sense". That is patently false. The "usual definition of speed" is v=|dx/dt|, and nothing else. Speed is defined prior to any mention of coordinate transformations, and its definition does not induce a special coordinate transformation. It cannot, because both the speed and the spacetime coordinates from which it is derived are all defined from a single reference frame.

The only way you could possibly think that there is any kind of logical contradiction between the Lorentz transformation and the "usual definition of speed" is if you assume at the outset that the "usual definition of speed" includes--whether explicitly or by implication--some other coordinate transformation. In this case you seem to be assuming that the Galilean transformation is forced on us by the definition of speed. That is simply false, and it seems to be the source of your errors.

But the Galilei transformation is forced on us by the definition usual definition of 'speed' (or 'velocity'):
with the velocity of an object with coordinate x defined as v=dx/dt , you have for the difference between the coordinates of two objects x1 and x2
v1,2=d(x1-x2)/dt = dx1/dt -dx2/dt =v1-v2
which is nothing else but a vectorial velocity addition (i.e. a Galilei transformation).
It is therefore obvious that the problem with the 'velocity' of light lies with the usual definition of the velocity v=dx/dt, but not (as Einstein assumed) with the space and time coordinates x and t (which are, unlike v, fundamental but not derived quantities).
The only possible interpretation of the invariance of c is thus that the travel time of a light signal does only depend on the coordinates of source and observer at the time of emission but not on their velocity (see my webpage regarding the Speed of Light for more).

Note also that the 'relativistic' particle dynamics could be accounted for without using the Lorentz transformation if the static forces are assumed to be velocity dependent in an appropriate way (see my page A Newtonian Relativistic Electrodynamics).


Thomas

 

[Tom2]

As I said before, the vectorial velocity addition only apparently disappears due to Einstein's re-scaling of the space and time coordinates. It is very much like claiming that you stay on the same spot when making a number of steps to which you have assigned the length 0 by definition.

No. The vectorial velocity addition is never invoked and it is not implied by linear transformations. I've already explained this.

The motivation to treat the propagation of light in the same way as that of ordinary objects (i.e. involving a frame dependent velocity) is clear from the similarity of the Galilei and Lorentz transformation

x'=x-vt and
x'=gamma*(x-vt)

No. This is your hangup, not Einstein's. As I already said you are forcing the Galilean transformation into the derivation, and you're doing it for no reason at all.

Since the first equation is the limit of the second for small v (gamma=1 then) , one can hardly argue that there is a vectorial velocity addition implied with one but not the other.

No. Vectorial velocity addition only holds when c-->infinity. It quite obviously fails for any finite value of c.

But the Galilei transformation is forced on us by the definition usual definition of 'speed' (or 'velocity'):
with the velocity of an object with coordinate x defined as v=dx/dt , you have for the difference between the coordinates of two objects x1 and x2
v1,2=d(x1-x2)/dt = dx1/dt -dx2/dt =v1-v2

No. That gives you the relative separation of objects 1 and 2, as determined by a single observer. It does not give you the velocity of one object relative to the other. It's plainly obvious that you are concluding that the definition of speed implies vectorial velocity addition only because you are assuming that it does.

which is nothing else but a vectorial velocity addition (i.e. a Galilei transformation).

No. I already explained to you that x1, x2, and t are all assigned from the same reference frame. You don't have a coordinate transformation until you start referring to coordinates from a second reference frame.

You simply do not understand what a coordinate transformation is.

 

[Tom2]

So what exactly is spacetime? Are you claiming that mass only curves a coordinate system that doesn't represent anything physical?

"What spacetime is" is an open question. Some hold the Newtonian view that it has a seperate existence apart from matter and energy. Others share the point of view of Leibniz that spacetime is defined only by the spatial and temporal relations that events have with one another. Whatever your interpretation of spacetime, your coordinate system is an abstract, mathematical construct that gets superimposed on spacetime. It's how an observer keeps track of events. Spatial measurements were always known to be coordinate system dependent. What SR adds to this is that time is also coordinate system dependent.

OK, let me get more specific. You have two clocks and an observer that are all stationairy relative to each other, but the observer and one clock are stationairy in spacetime, while the other clock is moving relative to spacetime. Relative to the observer, will the clock that is moving through spacetime function differently than the clock that is stationairy in spacetime?

I have no idea of what it means to be moving or stationary relative to spacetime, but I do know that if three frames are stationary relative to each other, then there is no sense in which they are moving relative to each other. That follows from logical transitivity.

One small unrelated point.

Prosoothus, to 2inquisitive: Actually, you're wrong. Relativity states that time can be faster or slower in another frame.

Dale: No, Prosoothus. Look at the Lorentz transform. The gamma factor is always greater than or equal to 1 so moving clocks never run faster than stationary clocks in SR.

I don't know if this is what you were talking about, but SR does indeed state that for any temporal interval Δt, the transformed interval Δt' can be greater than, smaller than, or equal to Δt. Yes, gamma is always greater than or equal to 1 but if we look at the Lorentz transform,

Δt'=γ(Δt-vΔx/c²),

we can see that Δx is an independent variable and it can be chosen so that Δt' greater than, less than, or equal to Δt.

 

[CANGAS]

Stereotypically, the Special Relativists use SR to prove SR.

 

[Pete]

Looks like a non-sequitur.

Were you responding to any post in particular, CANGAS, or did you simply feel the urge to perpetuate the crackpot's stereotype of the mainstream?

 

[CANGAS]

Pete, do you ever make a statement, or, do you always vent your urge to query?

Do you see me only ask questions?

Huh?

 

[DaleSpam]

Stereotypically, the Special Relativists use SR to prove SR.

Stereotypically, the crackpots do not want to use SR to determine what SR predicts

-Dale

 

[Tom2]

Stereotypically, the Special Relativists use SR to prove SR.

Stereotypically, the troll is pretending like he understands what is going on by making a non-specific comment about all the things that have been posted here. This discussion is about the deductive validity of SR. When you check the validity of a formal system, you're supposed to look within the system itself. This is basic logic.

 

[Tom2]

2inquistive:

Below is another post of mine that got lost in the shuffle. Could you answer it when you get a chance? Thanks.

I think I'm getting there:

I am talking about Special vs. Galilean Relativity.

You are talking about the absolute (a la Newton) vs. the relational (a la Leibniz) view of time.

Is that accurate?

 

[2inquisitive]

2inquistive:

Below is another post of mine that got lost in the shuffle. Could you answer it when you get a chance? Thanks.

To tell you the truth, I am not familar with the Leibnez view of time. I will have to try to show what I am speaking of in another way.

Start with two objects in an otherwise empty universe, A and B. The two objects can detect no relative velocity between them, so time is assumed to beat at the same rate in each of the inertial frames, identical in A and B.

Now, fill in the rest of our known universe. Let object A be an observer standing on Earth's equator. Let object B be a geosynchroneous satellite direct over observer A. There is no RELATIVE velocity between the observer and the satellite, but both are moving through spacetime (similar to an ether). To get the time dilation due to velocity through spacetime, we can use a geocentric reference frame, a frame with the coordinates centered at the Earth's core and not rotating with respect to the fixed stars. We then see that observer A standing on Earth's equator is moving through spacetime at a different velocity than the geostationary satellite (B). Time does not beat at the same rate at each location, the clock at B's location beats slower than the clock at A's location and BOTH clocks beat slower than a clock at the center of the Earth. The clock offsets could be calculated once the velocity of A relative to the geocentric frame and the velocity of B relative to the geocentric frame were known. But that STILL would not give the 'real world' offset of the clocks, the equations are lacking the 'sperm' I used as an analogy for gravity.

We next have to consider gravitational influence on the clocks. Clock B is higher in the gravity well and beats faster than A because of this. Both the clocks will beat SLOWER than a clock in 'flat' spacetime, a clock not influenced by gravity.

Now, think about how flat spacetime could be modelled in this gedankin. The MOTIONS of clock A and B were relative to an imaginary flat spacetime. The clock dilations due to gravity were relative to this same imaginary flat spacetime. Whether anyone wanted to call this flat spacetime an 'ether' frame is open to interpretation. Are my thoughts too far removed from actual physics?

 

[DaleSpam]

Start with two objects in an otherwise empty universe, A and B. The two objects can detect no relative velocity between them, so time is assumed to beat at the same rate in each of the inertial frames, identical in A and B.

Now, fill in the rest of our known universe. Let object A be an observer standing on Earth's equator. Let object B be a geosynchroneous satellite direct over observer A. There is no RELATIVE velocity between the observer and the satellite, but both are moving through spacetime (similar to an ether).

This is kind of interesting. But I think it would be more interesting if we left the rest of the universe empty. I mean, if there is the rest of the universe then you can say that the earth is rotating and the satellite is orbiting around it. But what if you don't have the rest of the universe? Then it appears that there really is no motion at all involved in a geosynchronous satellite over an otherwise isolated earth. I think you would have to postulate centrifugal force as a fundamental force of the universe or something.

Interesting.

-Dale

 

[MacM]

I think you would have to postulate centripetal force as a fundamental force of the universe or something.

Interesting.

-Dale

WHAT? You now want to makes GR's ficticious force a real force. LOL.

 

[DaleSpam]

Oops, for once you are right. I should have said "centrifugal" force. I will correct the original comment.

-Dale

 

[Pete]

Start with two objects in an otherwise empty universe, A and B. The two objects can detect no relative velocity between them, so time is assumed to beat at the same rate in each of the inertial frames, identical in A and B.

Now, fill in the rest of our known universe. Let object A be an observer standing on Earth's equator. Let object B be a geosynchroneous satellite direct over observer A.

It doesn't follow.
A and B can find out whether they are rotating or not by experimenting with a precision gyroscope or sagnac interferometer. This means that they can potentially stop rotating, and then see that the other is moving relative to them.

Furthermore, there must be some force or pseudo-force always acting on both A and B toward the "center of the Earth" point, and a measurable counter-force keeping A away from that point, which breaks the symmetry between the two.

So according to SR and GR, it should not be expected that the clocks of A and B will tick at the same rate.

 

[2inquisitive]

Pete, let's try another example.
Again, assume there are two objects in an empty universe, A and B. These two objects detect that there IS relative velocity between them, say they are approaching each other at some velocity relative to their individual rest frames. Each will assume the others' clock is beating slower than the one in his own rest frame.
Now, let object A be a satellite in the GPS constellation. Let object B also be a satellite in the GPS constellation, a different one of course. In either a geocentric frame of reference, OR a flat spacetime frame of reference, the GPS satellites will be travelling at identical speeds relative to either of these frames. Their clocks will beat at the same rate. Add 22 more satellites all travelling at the same speed relative to either of these frames of reference, and you have a synchronized satellite constellation, all their clocks beating at the same rate. Sound familiar?

 

[Pete]

Again, assume there are two objects in an empty universe, A and B. These two objects detect that there IS relative velocity between them, say they are approaching each other at some velocity relative to their individual rest frames. Each will assume the others' clock is beating slower than the one in his own rest frame.

Neither needs to assume anything. It's measurements that count.

If A and B are travelling similar circular paths so that they meet over and over again, each will measure the other's clock to run on average at the same speed as their own.

 

[2inquisitive]

Neither needs to assume anything. It's measurements that count.

If A and B are travelling similar circular paths so that they meet over and over again, each will measure the other's clock to run on average at the same speed as their own.

'The moving clock runs slow' Pete. How could an observer on satellite A ever measure B's clock beating faster than his, so as to 'average' out? Vice versa for B measuring A's clock from his rest frame.

 

[Pete]

The clocks are not SR inertial.

This is basic twin paradox stuff.
Just like Terrence's clock runs really fast for Stella during her turnaround, so A's clock runs fast according to B (and vice versa) at times during their orbit.

You should learn to do the maths, 2inq. It makes things so muich easier to see.

 

[MacM]

Neither needs to assume anything. It's measurements that count.

If A and B are travelling similar circular paths so that they meet over and over again, each will measure the other's clock to run on average at the same speed as their own.

Dang. I didn't realize that GPS would be so simple. Quick e-mail the Feds and tell them to stop prelaunch calibrating the satellites +7.2us/day for relative velocity to the ECI.

 

[DaleSpam]

Pete is right MacM.

But go ahead and send that e-mail. I am sure that everyone will be happy to have your GPS system that would need to average over 12 hours for a given satellite to get a correct location.

-Dale