Undergrad Change of special relativity formulas?

[Tahmeed]

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. 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?

 

[Dale]

Lorentz Transformation, Length contraction, Time dilation

None of these would change.

 

[PeroK]

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. 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?

There is a big difference in many cases between a measurement and an observation. A measurement is a calculation based on observations. If there is no local observer of an event in your reference frame, then it is practically more difficult (it may even be impossible) to measure the time and place of that event. But, the difficulty of measuring an event does not change its spacetime coordinates! You would of course have to work out how far the event was and then take into account the time lag for light (assuming there was light from the event).to reach you.

 

[Herman Trivilino]

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. 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?

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.

 

[Ibix]

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?

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.

 

[Nugatory]

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.

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]

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?

 

[Ibix]

If you factor in the lightspeed delay, you won't see straight length contraction, no. If you search this forum for "Terrel rotation" you might find some animations of Terrel rotation (which is what you see) that someone did a few years ago.

 

[m4r35n357]

If you factor in the lightspeed delay, you won't see straight length contraction, no. If you search this forum for "Terrel rotation" you might find some animations of Terrel rotation (which is what you see) that someone did a few years ago.

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.

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).

 

[Bartolomeo]

I won't see any sort of length contraction?

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 $\alpha =\arcsin v/c$
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.

 

[Ibix]

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.

 

[Android Neox]

In Relativity there is no distinction between reference frames and observers. In Relativity (and all physical science) observed (measured) reality is reality.