"Light is frame-dependent in PF, but constant in SR"

Get a clue. In the local coordinate frame and the local proper frame the measurement has been invariant in every empirical test. That frame has the same definition regardless where you make the measurement. It's sometime referred to as the Laboratory frame. The place where we conduct experiments. You think we have to go everywhere to prove it's a constant? That's nonsense and the biggest drawback to you learning any of this science. I'm not providing you with anything since I don't care whether you understand this or don't understand this. I don' know what you mean about testing where there's time dilation. Time and distance measurements in the local coordinate and local proper frames are invariant also. The local inertial frame is LOCAL. IE Local. So if we go far away and make the measurement we're actually making a local measurement. IE where the measurement actually occurs. So how about you telling me events in spacetime are not invariant. The local measurement is an event in spacetime. It's an invariant. If it was frame dependent then we would have noticed it by now. I took you off my ignore list but you just piss me off with the nonsense. Most real physicists will say nothing is absolutely proven but don't let that get in the way of doing science.

 

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That's what I've shown in a specific example with numbers, and you're incapable of seeing it in such a simple form, with simple math. As usual you only make broad known scientific statements (most of us already knew) and you cannot demonstrate any comprehension; never being able to point to specifics, making claims directly counter to what was shown, inferring something is incorrect yet unable to point to it, and you do not make an argument demonstrating any knowledge whatsoever of the post you refer to. You've proven countless times on Physforums your expertise is to paraphrase what you've read, or a straight out copy-paste it; I should have known your scientific knowledge and contributions to any discussion stops at being a source of reference material.

Maxilla

P.S. To other posters: I am open, and wish to review any mistakes or flaws I may have made in my post. I just ask you make an argument specifying the error(s) and demonstrate you have some comprehension of what was posted in your reply.

 

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Why did you post this as the reply to me? You know I am discussing SR not GR, and you should know at the velocities and distance I discussed (.98c) the Earth’s gravity has a negligible effect on the results and they can be ignored. It might look impressive to some, but I know better, this is a deflection, unnecessary, and irrelevant to the issues in the OP. Either specifically speak to that content, and stop pretending you know what I was talking about, or trying to impress people. You have yet to show you understood a single thing I posted; even what you implied I said in your criticism was incorrect and directly in contradiction to what I showed.

Maxila

 

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You are essentially correct, Maxila. Something similar came up on the other thread, it's worth repeating: See the simple inference of time dilation due to relative velocity. Imagine that you're zooming across the sky carrying a parallel-mirror light clock. I'm on Earth watching through my gedanken telescope, and I can see the light moving back and forth between your mirrors. According to me it's moving zigzag style, like this /\/\/\/\/\/\/\ at 299,792,458 m/s. According to you it's moving vertically up and down like this || at 299,792,458 m/s. However when I pan to follow you across the sky I too now see the light moving vertically up and down like this ||, and I see it moving slower. I am effectively looking into your frame from the outside. I can record everything I see on tape and put it up on a split screen. You get back, and peer over my shoulder. You hopefully now see the big picture. The light was moving slower in your frame, but locally you measured the same old value because within your frame all electromagnetic processes in your brain body and clocks were moving slower too. Sadly some people reject this without counterargument because they prefer to cling to "just-so magick".

 

There are two problems with this overly simplistic fantasy explanation of time dilation.

1) Symmetry. There is no way to identify the absolute frame of reference that this explanation presupposes. Anyone can set up this particular video experiment. (1a. This little video experiment will not actually work unless the video is put together later and then time adjusted.)

2) There is more to temporal events than merely photons. The time dilation also works for events like nuclear decay that do not involve photons. The timing of nuclear events that does not involve photons also changes between frames. Thus one cannot merely appeal to photons.

 

I’ve been following that thread and I think where you might lose people is they have the impression you are saying light is not an invariant value in every inertial frame? Experiments have proven this and I think you’d agree? In my OP I showed this for the two frames of reference in that example. What people seem not to be able to follow is that while those values are constant in every frame in terms of units, the units themselves cannot be equal, which can be demonstrated with very simple math.

In my OP (most of the data came from department of Physics at Georgia State University), from the ground the distance x is seen to the muon is 10km, from the muon that distance is seen as 2km, the time t from the ground for light to travel to the muon using units of.3km per microsecond is 33.33µs, From the muon frame that time is seen as 6.66 µs where the relative velocity of the muon to the ground frame is .98c and the gamma factor is 5. Let me make a chart of this:

[table="width: 500, align: center"]
[tr]
[td][/td]
[td]Distance[/td]
[td]Time(c@.3km/µs)[/td]
[td]Light Speed[/td]
[/tr]
[tr]
[td]Ground Frame[/td]
[td]10km[/td]
[td]33.33µs [/td]
[td].3km/µs[/td]
[/tr]
[tr]
[td]Muon Frame[/td]
[td]2km[/td]
[td]6.66µs[/td]
[td].3km/µs[/td]
[/tr]
[/table]

I think you’d agree both frames see c @ .3km/µs and that value would be seen constant for any inertial frame. What the numbers also show us is that the value for x and t in the muon frame relative to the ground frame is x/5 and t/5 to the grounds x and t (the inverse of the gamma value 5). If the relative motion of c in the muon frame were not also c/5 to the ground, than muon must observe c travel 10km and 33.33µs the same as the ground, and it does not. If the observed values of x and t for the muon are 1/5 the grounds than it is implicit the muon units of 3km/µs are equivalent to 1/5 the grounds also.

This relationship is 10km/2km = 33.33µs/6.66µs = .3km per µs/(.3km per µs/5) or we can say 2km/6.66µs = c = 10km/33.33µs with the constant value for c being .3km/µs. The only way c can remain invariant in each frame is for its motion to remain constant to how each frame values x and t. This is in agreement with SR evidence, implicit in the math, and consistent in that empirically clocks measure motion to derive time.

Maxila

 

Farsight, this is brilliant. I must point out that we can't quite make any absolute claims here because the rocket ship occupant could make the same conclusions about the Earth lab light clock. That being said, I admire your creative tenacity in the search for truth.

 

You need to knock off all of the pseudoscientific nonsense in the Physics & Math forum. There are fora specifically designated for this crap. There is nothing uncivil about stating facts, and if you cannot recognize that this thread has become a lightning rod for cranks then you probably cannot identify pseudoscience.

You are trying to directly compare the muon and ground frames without using the Lorentz transform necessary to do so. Your comparison is simply physically invalid. But no doubt you will continue your nonsense regardless of any amount of reason, so I will be counting the days until this thread is closed/moved.

 

I'd sincerely prefer to remain civil and not appear insulting, however per your remarks; your statement above demonstrates you don't have a basic knowledge of SR, gamma is the Lorentz factor (see http://en.wikipedia.org/wiki/Lorentz_factor ). I stated the gamma factor for the transform was 5 in my examples. Learn a little SR and you're less likely to look foolish when trying to discuss it, you likely don't relaize just how ludicrous that comments is.

Maxila

 

The invalid physical comparison which leads you to write "c = (.3km/1 µs)/5 for the muon" is what I am talking about. If you understand SR then between this and claims that c is frame dependent, you should not be the least bit surprised with the reaction you get.

 

Let's discuss it, how is it invalid? You don't expect a reader to take your word for it since you've not demonstrated any knowledge thus far? This is a science discussion, if you wish to make an argument with relevant facts in a scientific manner, do so. If you continue to spew a few implicit words while not demonstrating actual knowledge, I won't waste time replying to anymore of that nonsense.\

Those are your two quotes above, it's clear you said, I did not do a Lorentz transform for the ground frame and Muon frame. After I explained to you what gamma was, you made up something new.

I don’t care, if you can start discussing science and stop *pretending you know what you are talking about (*see Syne's next comment below).

Maxila

 

No, it is your claim to support, so do your own homework. There is copious evidence that c is invariant in all inertial frames, including that of the muon. So quit playing the typical pseudoscience game of shifting the burden of proof.

 

How did you arrive at 5?

When (Gamma, as the Lorentz factor) \(\gamma = \frac{1}{sqrt(1-v^2/c^2)}\)

 

Perhaps you should bother to familiarize yourself with this forum and its regular, long-term posters. You are new here, and you are demanding what every hack that posts here demands, that someone else take the time they did not bother to take to show where they were wrong. This is called confirmation bias, where you simply stop studying once you think what you believe has been verified. Most of us know the uselessness of trying to overcome such bias, especially when such a new poster is already going on the offensive (entrenched beliefs). That usually does not bode well for your time here.

I understand SR enough to instantly recognize crap when I see it, and any result that claims a different speed of light is just that, as it directly violates one of the postulates of SR (and cannot possibly be arrived at using a Lorentz transform). You would know this if you knew SR half as well as you so vehemently pretend to. And no, your trollish baiting is not going to make me feel that I have anything to prove to you, so you can give it rest.

 

As I mentioned in my post, I used most of the data already calculated from the Georgia State University Physics department (a link was provided), however I’d be happy to calculate it for you for verification.

γ = 1/(1-v^2/c^2)^1/2 Where c=1 and the velocity is .98c:

γ = 1/(1-.98^2/1^2)^1/2

γ = 1/(1-.9604/1)^1/2

γ = 1/(1-.9604)^1/2

γ = 1/(.0396)^1/2

γ = 1/.19899

γ = 5.0253 Rounded to:

γ = 5

Maxila

P.S. @Syne: To the intelligent and knowledgeable reader, your remarks speak from themselves. I no longer need to point out you are a pretender who does not know what he is talking about, since you are so adept at demonstrating that yourself.

 

gamma worked out to 5.025189076 before rounding, but the over all answer is correct.

 

You haven't done anything with the maths, other than the most trivial notion of what speed is. You could have at least applied the notion of Lorentz transforms properly. Too many people who are not familiar with special relativity just trot out \(t' = \gamma t\) and \(l' = \gamma l\) therefore \(l'/t' = l/t\). That is a superficial but flawed description of what the Lorentz transformations do. It is what people who only read wordy summaries of special relativity think relative motion does within special relativity. It doesn't take into account different spatial positions or any kind of vector formalism.

A more technical and precise formalisation of the light speed invariance is that under a Lorentz transform null geodesics are mapped to null geodesics. A null 4-vector is one which satisfies \(v^{\mu}v_{\mu} = 0\). A null curve is a path whose tangent vector is always null. A null geodesic is a geodesic which is a null curve. Light moves along null geodesics in all inertial frames. Non-inertial frames or equivalently curved space-time are not bound by this condition, hence why you can construct scenarios where someone might see a photon moving at some speed other than c, as Bruce has already mentioned.

The construction of the most general transforms allowed then follow by the null condition, \(v^{\mu}v_{\mu} = 0\), since \(v^{\mu}v_{\mu} = \eta_{\mu\nu}v^{\mu}v^{\nu}\). Applying a Lorentz transform \(v^{\mu} \to \Lambda^{\mu}_{\rho}v^{\rho}\) then gives the null invariance condition in the form \(\Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\lambda}\eta_{\mu\nu} = \eta_{\rho\lambda}\). Actually the null condition isn't enough, the invariance must apply to all 4-vector norms, \(v^{\mu}v_{\mu} \to (v')^{\mu}(v')_{\mu} = v^{\mu}v_{\mu}\).

There's a difference between stating something as simply as possible and just being wrong.

Wind your neck in.....

As I said, your attempt at applying Lorentz transforms is flawed, in that you just multiply top and bottom by the Lorentz factor, which is not how Lorentz transforms work. It may be very tempting to do that, to just cut a corner when it comes to time and space dilation but that isn't how it works. There's a reason introductory special relativity textbook make comments about coordinates and how their origins are taken to intersect etc. It is because a proper 1+1 dimensional Lorentz boost looks like \((t,x) \to \big( \gamma(t-vx) , \gamma(x-vt) \big)\) (in c=1 units). Clearly \(\frac{x}{t} \neq \frac{x'}{t'}\).

In fact the tendency for people to describe Lorentz transforms as acting on x and t is itself deeply flawed and open to 'abuse'. This is because x,t are coordinates, they define locations within a space-time. The Lorentz transforms are not defined by their action on coordinates of a manifold but by their action on the tangent space of a manifold. Special relativity allows for this abuse in many cases because the manifold is flat and often inertial motion is what we're interested in. But if we push things too far and forget we've made this simplification problems can arise. To be more specific recall what I just gave as the definition for a Lorentz transform, 4-vector norms must be invariant. The Lorentz transforms act on the vectors, not on the location of the vectors. A 4-vector v expanded in a tangent space basis \(\partial_{a}\) is \(v = v^{a}\partial_{a}\). We define the metric components by \(\eta_{ab} = \eta(\partial_{a},\partial_{b})\) and so by linearity of the metric \(\eta(v,w) = v^{a}w^{b}\eta(\partial_{a},\partial_{b}) = v^{a}w^{b}\eta_{ab}\). The tangent space basis \(\partial_{a}\) is short for \(\frac{\partial}{\partial x^{a}}\) where \(x^{a}\) are the coordinates so we can associate the Lorentz transform on \(\partial_{a}\) with a transform on \(x^{a}\) and it comes out to be the same as how the \(v^{a}\) components transform but when you start messing with this stuff in anything but the most trivial ways things rapidly stop simplifying to the forms non-physicists are familiar with.

When you can demonstrate any of your claims are physically viable and not just "Because Farsight says so" then you get to make that statement without being called a dishonest hypocrite. Until then you're a dishonest hypocrite for saying such a thing.

Mmmmm, delicious irony given your failure to do any of the Lorentz transform stuff properly. You are giving the impression you believe yourself familiar with special relativity, given how much you're trying to deride others for their supposed lack, yet you opened a thread with comments like the following ones :

That's a lazy short cut which is a classic mistake people learning relativity for the first time often make. Yes, you can view it either as from the ground's point of view where the muons decay slower than usual or from the muons point of view where the atmosphere is lower in height, but the specifics of where various objects, such as muons, see one another and the ground involves the more general Lorentz transform formalism I just went through.

That is a complete non-sequitor. You don't just get to throw in a factor of gamma randomly like that, though I can see why you did it. A larger fraction of muons reach the Earth's surface than Newtonian mechanics predicts, fine. The fraction is a frame independent things. If you're considering time and space dilations then the Earth observer might want to say "The muons' clock ticks slower by a factor of \(\gamma\) so more time" while the muons will say "The Earth's atmosphere is less deep by a factor of \(\gamma\) so less distance". So if you're ignorant to how SR works on a quantitative level and all you know about Lorentz transforms is the gamma factor you might say "Well now we've gone from \(\frac{L}{T}\) to \(\frac{1}{\gamma} \frac{L}{T}\)!" and by rearranging \(\frac{1}{\gamma} \frac{L}{T} = (\frac{L}{\gamma}) \frac{1}{T} = \frac{1}{\gamma T} \frac{L}{1}\) we have the two points of view I just described but you aren't comparing two different frames connected by a Lorentz transform, you're connecting the Newtonian concept with a somewhat informal concept in special relativity.

To see what two different inertial frames have to say about the dynamics of something you have to properly define the dynamics in a particular set of coordinates (ie one of the inertial frames) and then apply the appropriate Lorentz transform and then look at the resultant new view of the dynamics. If Frame 1 sees a photon moving at light speed (ie on a null geodesic) then Frame 2 will also see it moving along a null geodesic, as I formalised earlier in this post. That is how to do it properly, none of this 'palming a card' by failing to apply Lorentz transforms properly and comparing two quite different conceptual models.

If you want to discuss relativity then great. However, if you don't know how to do special relativity please refrain from pretending you do and calling into question other people's knowledge when you're showing you're not exactly a whiz at it.

 

See the appropriately scathing reply from AlphaNumeric.

 

There’s a lot here, I’d like to go over smaller pieces in a few posts over more time.

I learned to use that shortcut because it results in the correct answer. I mentioned in the OP most of the data (including those transformed distances and times) came from the Georgia University Physics dept. site (link was included), I did calcualte gamma when asked above. I used the shortcut to demonstrate a relationship of each frames space time relative to the other. If I had calculated the relative distance or time using a Lorentz transform it would yield the same result.

I understand that all coordinates are transformed relative to each frames space-time, yet I am uncertain if, or how, you might be indicating it alters the data I discussed?

Inherent for a value of c is x/t, as a change in distance (x) per time (t). Where you lose me in the statement above is anytime we state x and t have different relative values as coordinates, than I can’t see how those values as a change in distance (x), per time (t), inherent in c cannot also change proportionally, or it would no longer be measured invariant for the frames coordinate values of x and t?

Even if I look at a time dimension as ct, it implies c as a change in distance per time is a constant relative to a frames value of x, t (the value of c as being (x/t)*t)? I don’t see that disagreeing with SR, rather just being implicit within SR, in that space coordinates and motion are relative to a frame, and c is a constant value for motion relative to the frame’s view of space (t, x, y, z)?

Fair enough, I did not overstate my SR knowledge in my OP, however I did imply otherwise in latter ones. I was attempting to goad a useful and constructive reply such as yours, instead of meaningless inference and innuendo. I know of, and respect your credentials. You given me plenty to consider and I look forward to digesting it.

Maxila

 

Baiting people with inflammatory posts is typically called trolling.