#### DeltaForce

Gold Member
Summary
I have a question about simultaneity.
The famous example demonstrating that simultaneity is an non-invariant variable would be the lighting bolts striking a fast moving train. Or a projector at a mid-point shooting off two beams at the same time to receivers on the opposite ends (for the person at rest and not on the train's reference frame)

My question is (and this may sound stupid): Is simultaneity an non-invariant variable only when the "signal" fired off at the mid point is traveling at the speed of light. As the speed of light is absolute and constant throughout all inertial reference frames. So.... anything less than the speed of light, the effects of simultaneity an non-invariant variable wouldn't be shown?

If you don't understand what I'm trying to say, it's okay. I'm also not totally sure how simultaneity works in general. My head is all jumbled up on the inside so I don't really know what the hell I'm talking about. But this simultaneity shenanigans is really weighing on my mind right now, I really need to get it out of the system.

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#### Orodruin

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No, in fact the relativity of simultaneity has nothing to do with signals or speeds, the example with the lightning bolts is just a demonstration of the concept. When we talk about simultaneity what we talk about is the simultaneity of two events, which are points in spacetime. Two events that are simultaneous in a given reference frame $S$ by definition have the same time coordinate $t$ and different space coordinates $x_1$ and $x_2$, respectively. If you make a Lorentz boost to a frame $S'$ moving with speed $v$ relative to $S$, you will find that
$$t'_1 = \gamma(t - vx_1/c^2), \quad t'_2 = \gamma(t-vx_2/c^2) \quad \Longrightarrow \quad (t_2 - t_1) = \frac{v(x_1-x_2)}{c^2}.$$
Therefore, the events do not occur at the same $t'$ coordinate and therefore are not simultaneous in $S'$.

#### Ibix

So.... anything less than the speed of light, the effects of simultaneity an non-invariant variable wouldn't be shown?
No. It's just harder to show with signals that don't travel at the speed of light.

I suspect you are imagining using a pair of guns with some muzzle velocity $v$ instead of flashlamps at the centre of the carriage. In the frame where the guns are moving at speed $u$ the bullets have velocities $u+v$ and $u-v$ and they always strike the carriage ends simultaneously, right? Unfortunately, this is wrong. The bullets actually have velocity $(u+v)/(1+uv/c^2)$ and $(u-v)/(1-uv/c^2)$, so they strike non-simultaneously. The reason that you don't notice at every day speeds is that $uv/c^2$ is tiny, so the difference in simultaneity is well below your ability to detect (it makes a difference to about the fourteenth decimal place).

You don't need to do any of that if you do the experiment with light pulses because their speed is always $c$ by definition in this theory.

I strongly advise looking up how to draw Minkowski diagrams if you are confused about simultaneity. They are a really neat way to visualise relativity.

#### pervect

Staff Emeritus
Summary: I have a question about simultaneity.

The famous example demonstrating that simultaneity is an non-invariant variable would be the lighting bolts striking a fast moving train. Or a projector at a mid-point shooting off two beams at the same time to receivers on the opposite ends (for the person at rest and not on the train's reference frame)

My question is (and this may sound stupid): Is simultaneity an non-invariant variable only when the "signal" fired off at the mid point is traveling at the speed of light. As the speed of light is absolute and constant throughout all inertial reference frames. So.... anything less than the speed of light, the effects of simultaneity an non-invariant variable wouldn't be shown?

If you don't understand what I'm trying to say, it's okay. I'm also not totally sure how simultaneity works in general. My head is all jumbled up on the inside so I don't really know what the hell I'm talking about. But this simultaneity shenanigans is really weighing on my mind right now, I really need to get it out of the system.
No. For a specific example, consider firing off electron beam, using the same midpoint formulation of simultaneity.

One says that electron beams of a known energy in an inertial frame travel at "the same velocity", regardless of direction. This is because the universe is "isotropic". We don't think, for instance, that identical electron beams pointed north should have a different velocity than ones pointed south. "Identical" can be quantified as "having the same energy" in this context, because the energy of the electrons in the beam controls the velocity of the beam.

The electron the share the same notion of simultaneity that light beams do, but they are not light and do not travel at light speed. This shouldn't be too surprising, because , while the beams don't travel at light speed for any finite energy, in the limit of very large energies they approach the speed of light.

For a reference on this point, which I think is very fundamental and interesting, see for instance the video "The Ultimate Speed", or the peer-reviewed paper written about it.

Really, the main reason to use light is that it's convenient, it always travels at "c" regardless of energy.

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