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Tahmeed

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In summary, the Lorentz transforms do not answer the question if an object is stationary or moving relative to a stationary observer.

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Tahmeed

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Dale

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None of these would change.Tahmeed said:Lorentz Transformation, Length contraction, Time dilation

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Tahmeed said:

There is a big difference in many cases between a

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Mister T

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Tahmeed said:

A derivation is just one way of demonstrating the validity of a relation. You may be able to come up with several different ways to derive the same relation, but changing the way you derive a relation doesn't have to change the relation itself, or the validity of that relation.

In this case you could station mechanical sentries at different locations. Those produce records of the position and time of each event. Those records can then be used in the same way as the observers. Another way would be a signal generator at each location that sends a signal to the one observer. Knowing the speed of the signal propagation and its distance away the observer can deduce the position and time of each event.

A good derivation simply takes the simpler, rather than the more complex, way to derive the result. It's a teaching tool. It doesn't change the physics.

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Ibix

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It's not clear to me what you are asking. Are you trying to ask something like: if I take a photo (with a high speed camera) of a moving object, will I see it length contracted? What will I see if not?Tahmeed said:But what if there is only one observer in a certain frame who is some d distance away from the event point? How should we change our equations?

If so, the answer is that you will not see length contraction in general. But that isn't because we need to change the Lorentz transforms. It's because the Lorentz transforms do not answer that question directly. You need to factor in the light speed delay.

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Nugatory

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That assumption is not part of the derivation. It's only there to help new students understand what it means to assign coordinates to spatially separated events; often derivations in more advanced presentations don't include it.Tahmeed said:We know that while deriving the special theory of relativity formulas like Lorentz Transformation, Length contraction, Time dilation etc, we assume that there is an observer at each point of the space in a certain frame and all these observers contain synchronized clocks and scales.

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Tahmeed

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Ibix said:It's not clear to me what you are asking. Are you trying to ask something like: if I take a photo (with a high speed camera) of a moving object, will I see it length contracted? What will I see if not?

If so, the answer is that you will not see length contraction in general. But that isn't because we need to change the Lorentz transforms. It's because the Lorentz transforms do not answer that question directly. You need to factor in the light speed delay.

I won't see any sort of length contraction?

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Ibix

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m4r35n357

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I'd say Terrell Rotation is not really the equivalent of length contraction here. It is merely an optical illusion caused by aberration in which things you would see from the side when at rest are "projected" further in front of you as you go faster, so they appear to be turning inside-out.Ibix said:

IMO the nearest equivalent to length contraction would be the Doppler effect, which actually stretches out things you are moving towards and only contracts the view to the rear.

I made these videos a while ago, if anyone is interested in what you can "see" at high speeds. I use quotes because I don't think anyone can really see something moving at such relative velocities (also the headlight effect would wash out the forward view and the rear would just go dark).

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Bartolomeo

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If the sphere (like in Terrell's article) is semi-transparent and a measuring rod is hidden inside the sphere, on the picture you will see that the rod is gamma times contracted, but the sphere is rotated at the angle ## \alpha =\arcsin v/c##Tahmeed said:I won't see any sort of length contraction?

Rays of light from the ends of the rod to the aperture and from the aperture to the film to form two similar triangles. They connect points of flat object and corresponding points on the film and aperture.

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Ibix

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OK - let's say you are stationary at x=0, y=Y. A pointlike object is moving along the x-axis at velocity v, such that ##x=vt+\alpha##, where ##\alpha## is some constant offset. Light emitted from the object at time ##t_e## reaches you at time ##t_r##. Pythagoras tells us that ##c^2(t_r-t_e)^2=Y^2+(vt_e+\alpha)^2##. That can easily be solved to get the emission time $$t_e=\frac{2v\alpha+c^2t_r-2\sqrt{(c\alpha+cvt_r)^2+(c^2-v^2)Y^2}}{2(c^2-v^2)}$$where I've chosen the negative sign.

Of course, we aren't interested in a pointlike object. We've got a measuring rod which has length ##L## in its rest frame. Hence, if it is moving at speed v in this frame and centered at ##x=vt##, then ##\alpha## varies along its length from ##-L/2\gamma## to ##L/2\gamma##. So the apparent rod length at time ##t_r## is ##v\left(t_e|_{\alpha=L/2\gamma}-t_e|_{\alpha=-L/2\gamma}\right)+L/\gamma##, which is$$l=v\frac{vL/\gamma-\sqrt{(cvt_r+cL/2\gamma)^2+(c^2-v^2)Y^2}+\sqrt{(cvt_r-cL/2\gamma)^2+(c^2-v^2)Y^2}}{c^2-v^2}+L/\gamma$$

*Edit: mangled on my phone screen. This may be a more useful form for small screen users:$$\begin {eqnarray*}l&=&v\frac {vL/\gamma-r_++r_-}{c^2-v^2}+L/\gamma\\r_\pm&=&\sqrt {(cvt_r\pm cL/2\gamma)^2+(c^2-v^2)Y^2}\end {eqnarray*}$$*

That does not look like a constant ##L/\gamma## to me. A quick plot for L=1, Y=10 and v=0.6 is below for ##-100<t_r<100##. Do check my algebra. It appears to have sensible limiting behaviour, but I literally did this on the back of an envelope.

Of course, we aren't interested in a pointlike object. We've got a measuring rod which has length ##L## in its rest frame. Hence, if it is moving at speed v in this frame and centered at ##x=vt##, then ##\alpha## varies along its length from ##-L/2\gamma## to ##L/2\gamma##. So the apparent rod length at time ##t_r## is ##v\left(t_e|_{\alpha=L/2\gamma}-t_e|_{\alpha=-L/2\gamma}\right)+L/\gamma##, which is$$l=v\frac{vL/\gamma-\sqrt{(cvt_r+cL/2\gamma)^2+(c^2-v^2)Y^2}+\sqrt{(cvt_r-cL/2\gamma)^2+(c^2-v^2)Y^2}}{c^2-v^2}+L/\gamma$$

That does not look like a constant ##L/\gamma## to me. A quick plot for L=1, Y=10 and v=0.6 is below for ##-100<t_r<100##. Do check my algebra. It appears to have sensible limiting behaviour, but I literally did this on the back of an envelope.

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Android Neox

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Special relativity is a theory developed by Albert Einstein in 1905, which explains the relationship between space and time. It states that the laws of physics are the same for all observers in uniform motion, and the speed of light in a vacuum is constant for all observers regardless of their relative motion.

Special relativity formulas take into account the effects of high speeds and the constancy of the speed of light, while classical mechanics formulas do not. This results in differences in the equations for time, length, and momentum between the two theories.

Some important special relativity formulas include the time dilation formula (t = t0 / √(1-v^2/c^2)), the length contraction formula (l = l0 * √(1-v^2/c^2)), and the relativistic momentum formula (p = m * v / √(1-v^2/c^2)). These formulas are used to calculate the effects of high speeds on time, length, and momentum.

Special relativity has greatly impacted our understanding of the universe by providing a more accurate and comprehensive explanation of the relationship between space and time. It has also led to the development of other important theories, such as general relativity and quantum mechanics, which have further enhanced our understanding of the universe.

Yes, special relativity formulas can be applied to everyday situations, although the effects are typically only noticeable at extremely high speeds. For example, the GPS system uses special relativity to make precise calculations for location and time, taking into account the effects of high speeds on the satellites in orbit.

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