What time dilation effect would occur when two objects near the speed of light approach each other?

Question

What time dilation effect would occur when two objects would approach each other at e.g. 90% the speed of light? For each of the objects the other object appears to be moving at 180% the speed of light relative to itself. In this case the formula for calculating the time dilation 'factor' can't be applied. How do I correctly calculate it?

Answer 1

So here's how you'd address the problem: I find a lot of confusion about Special Relativity is cleared up when you decide with respect to whom the velocities are being measured. I am going to assume that you are in the lab frame, observing two objects (call them $A$ and $B$) moving towards each other with a speed $0.9c$ each with respect to you. Let us first ask what velocity $A$ sees $B$ approaching it with.

You might naively say that:

$$\text{velocity of $B$ w.r.t $A$} = \text{velocity of $B$ w.r.t you} -\text{velocity of $A$ w.r.t you}$$

or mathematically

$$u' = u - v,$$

where $u'$ is the velocity that $A$ sees $B$ moving at, $u$ is the velocity you see $B$ moving at, and $v$ is the velocity of $A$ with respect to you. Using this you might conclude that $u' = -1.8c$ (as I believe you have). However, this is wrong, since you have used a "common-sense" velocity addition law that is no longer true in Special Relativity. You can derive the velocity addition law in Special Relativity quite simply from the Lorentz Transformations (see my answer here), and show that

$$u' = \frac{u - v}{1 - \frac{uv}{c^2}},$$

and so the actual velocity that $A$ will see $B$ moving at is $$u' = -\frac{1.8c}{1+0.9^2} = - 0.994475\, c,$$

which is quite close to the speed of light, but certainly not greater than it! (The negative sign symbolises that $B$ is moving towards $A$.)

Now, to answer your question, the object $A$ sees the object $B$ moving towards it at a speed of $|u'| = 0.994475 c < c$ and therefore sees the time dilate by the appropriate Lorentz factor associated with this speed $u'$, which happens to be $\approx 9.5$, or in other words, $A$ sees time passing nearly 10 times slower for $B$.

Answer 2

A spacetime diagram reveals what is really going on in special relativity.

Essentially, as others have said, one has to find the velocity $V_{BA}$ of Bob (worldline OB) with respect to Alice (worldline OA), given the velocities $V_{AL}$ and $V_{BL}$ with respect to the Lab (worldline OL).

Geometrically, one has to find the slope of Bob's worldline (shown in green) on Alice's spacetime diagram, leading to the Minkowski-right-triangle (shown in blue) with space and time components in Alice's frame. (The Galilean approach would be to use Alice's worldline and the dotted horizontal line as sides of the Galilean-right-triangle.)

For arithmetic simplicity, I'll use $V_{AL}=(3/5)c$ and $V_{BL}=(-3/5)c$ [since their corresponding Doppler factors, $2$ and $1/2$, are rational].
By drawing it on "rotated graph paper", we can more easily see the tickmarks along various worldlines. The key idea is that all "light-clock diamonds" (representing a tickmarks along any worldline) have the same area. (ref: my "Relativity on Rotated Graph Paper", American Journal of Physics 84, 344 (2016); https://doi.org/10.1119/1.4943251)

In special relativity, Alice measures the velocity of Bob using a convenient Minkowski-right triangle with sides [it turns out, using the method of my paper] (-15) and (17), which meets Bob's worldline after 8 diamonds (which is the hypotenuse of that Minkowski-right-triangle). So, $V_{BA}=(-15)/(17)$.
Time dilation is the ratio of the [timelike] adjacent leg (17) to the [timelike] hypotenuse (8): $\gamma_{BA}=(17)/(8)$.
Yes, $(17)^2-(-15)^2=(8)^2$.

(In galilean relativity, all measurements would use the diamonds in the Lab frame (the background grid) as the tickmarks. Alice would use a convenient triangle with the spacelike-leg parallel to the horizontal (representing absolute time in Galilean relativity), like the one through L. (Note that, in Galilean relativity, all timelike displacements from O that meet the horizontal line through L have magnitude 5.)
Alice would determine Bob's velocity as $V_{BA,gal}=(-6)/(5)$, and time dilation as $\gamma_{BA,gal}=(5)/(5)=1$ [no time dilation].)

For the case of $V_{AL}=(9/10)c$ and $V_{BL}=(-9/10)c$, one would have a triangle with legs (-180) and (181) and hypotenuse (19). So, $V_{BA}=(-180)/(181)$. And since $(181)^2-(-180)^2=(19)^2$, we have $\gamma_{BA}=(181)/(19)$.

One could use a velocity-composition formula to determine $V_{BA}$,
but it might be better to first put the spotlight on what is fundamentally happening (in a numerically simpler case) by drawing a spacetime diagram.

Answer 3

If two speed(say Alice and Bob) are given w.r.t to you ,and you are asking the relative speed as seen by Alice w.r.t Bob you have to apply relativistic addition for speed.
However if you ask what is the relative speed of Bob and Alice relative you , then certainly it's 0.9c+0.9c,
The general formula reads like this ,say for speed of alice relative to bob, $\frac{speed\: alice \: w.r.t\: other\: frame\: + speed\: of \:that\: frame\: w.r.t \:to\: Bob\:}{1+product\: of \:those\: speeds\; scaled\; by\; c}$
in your case this translates to ,$\frac{0.9c+0.9c}{1+0.9c*0.9c}=0.9945c$
Time dilation of alice w.r.t to bob can be calculated using this speed
Time dilation for alice and bob w.r.t you would simply depend on their individual speed (0.9 not 1.8c)
You have to choose a frame alice + bob is not a frame .