# The speed of tachyons

The other day I was wondering: When a tachyon is coming towards you faster than the speed of light, will you see it before it hits you? Then I thought of course not, since the light waves aren't traveling faster than the tachyon then how could you see it before it hits you? Now I thought today, if an tachyon is traveling away from you faster than the speed of light, would you see it?

If you fire a ball at an initial velocity of 20mph south out of a car that is going 50mph north, the final velocity of the ball would be 30mph north, is this also how light acts when the initial velocity of the object it is reflecting off is not equal to 0?

So in my case, if the speed of light were 100mph (dummy math) and a tachyon was traveling at 110mph north that means the light reflecting off the tachyon would be traveling at 10mph north, so then really would you be able to see it? More generally, how does relativistic addition of velocities work for tachyons?

update:

This question is a hypothetical question: IF tachyons exist, then what would happen? After a few hours of research I see why a usual massive object CAN'T travel faster than (or even reach) the speed of light, but this question is about tachyons.

• You may want to see enwp.org/Tachyon . You wil ony see it after it asses you, iirc then you'll see teo copies of it. Of course it's not possible for currently known particles and objects. Feb 26, 2012 at 6:22
• will the effect on an object with mass be proportional to the effect on a single particle?
– John
Feb 26, 2012 at 6:55
• You can't accelerate objects with mass to the speed of light. Conversely, anything that goes faster than light doesn't have normal mass. It has complex mass; and cannot be slowed down. Feb 26, 2012 at 7:06
• Currently faster than light stuff is all theoretical and controversial.. Feb 26, 2012 at 7:34
• i looked it up and i see where you're coming from, its not possible for an object with normal mass to travel faster than the speed of light, which really makes this a hypothetical question.
– John
Feb 26, 2012 at 7:34

If a tachyon starts from where you are and goes away at faster than the speed of light, you will see the photons it emits earlier than it actually departs. So you will see all these photons coming as if the tachyon were coming toward you at a speed slower than light, and then bang, the tachyon leaves. In fact, the faster it is going away, the slower it appears to be arriving.

EDIT: You can just tell this from a space-time diagram:

T  s   C
| /   /
|/   /    f
| s /    /
|/ /  f /
| /  / /
|/__/_/___X
|  / /
| / /
|/ /
| /
|/


Here, the time T axis is vertical and the space X axis is horizontal. Line C represents the speed of light. Photons move parallel to that line.

If something is moving away from you slower than light, it is a diagonal line falling in the slow (s) region. When it emits photons, they travel parallel to C, so each one arrives back at you at a later time. That's the normal behavior that you're used to.

If something is moving away from you faster than light, it is a diagonal line in the fast (f) region. When it emits photons, they travel parallel to C, and thus arrive back to you at a negative time, relative to when the object left you. In fact the faster it's moving (closer to horizontal) the earlier its photons will arrive (negative T). The slower it's moving (closer to C) the more its photons will appear to come all at once, just before it "departs".

• I gave you the answer because you went most in depth and you not only explained what you thought would happen, you supported it with the science behind it
– John
Feb 27, 2012 at 2:20

As the commenter already indicated, there is no way for masses to reach a speed faster than the speed of light, $c$ (or even as fast as $c$).

The analogy with the cars does not hold when you go to velocities close to $c$. Those velocities are called relativistic velocities. The Galilean transformations for low velocities do not hold any longer, instead, you should use the Lorentz transformations. This based on Einsteins special relativity theory.

For example, suppose you have an object at a distance of $1$ lightyear, moving with velocity $0.99c$ away from you. With classical physics you would calculate that the speed of the photon is $0.01c$ towards you, so it would take the light $100$ years to reach you.

One of the postulates of Einstein, is that the speed of light is always $c$, irrespective of the frame of reference. So, the light photon emitted will travel with $c$ towards you, so you would be able to see the object after one year (instead of $100$).

• thank you for answering, so you believe in the theory that something that is traveling at the speed of light cannot slow down, therefore in my case the light will always reach the person no matter the speed?
– John
Feb 26, 2012 at 17:27
• @John No, I have no idea what direction the photons would go to in the frame of the observer. It is too difficult for me to tell in this hypothetical situation. Feb 26, 2012 at 18:48

If you fire a beam of photons at an object receding away from you at a speed greater than the speed of light, your photons will never reach it to reflect off it

OTOH, if such an object emits photons, you should eventually be able to see the object as it was at the time the photon was emitted.

• You're not considering relativity here. Which you must consider. Feb 26, 2012 at 12:52
• @Manishearth can you elaborate please ? I dont see what relativity has anything to do with a emitted photon eventually reaching you Feb 26, 2012 at 13:36
• Speed of light is constant in all reference frames, you can't apply relative velocity. Whenever dealing with near-lightspeeds, you have to take relativity into account. You can get strange results otherwise. (you can get strange results with relativity as well, but th at s a different matter) Feb 26, 2012 at 13:48
• @Manishearth I understand that, but instead of talking generically about relativity can you make a concrete case. Assume that a photon is being emitted by a FTL object, if the object is moving away from you I understand that the photon will be red-shifted, but apart from that what has relativity got to do with it ? How does relativity modify this scenario ? Feb 26, 2012 at 13:54
• The redshift would be relativistic doppler shift. As with all relativistic stuff, you get crazy results if $v>c$ . The accepted answer is correct, see that for a more detailed explana tion. The emitted photon does not 'eventually' reach you, it already has reached you. Feb 27, 2012 at 2:28

The short answer is that no, light doesn't behave like a ball being thrown out of a moving car. The speed of light in a vacuum, $c$, is a universal constant, meaning that all light always travels at the same speed, independent of the speed of the thing that emits it. So you would be able to see your faster-than-light object moving away from you, because its light would be travelling towards you at $c$.

The slightly longer answer involves explaining that the speed of light is also independent of the speed of the person observing it. This seems impossible at first, but Einstein realised it was possible if you re-think the nature of space and time, and that's the basis of special relativity. Special relativity ultimately shows you that (i) it's impossible to accelerate an object any faster once it gets to the speed of light, and (ii) if objects could be created travelling faster than $c$ (thus avoiding the need to accelerate them), it would be possible to use them to send messages back in time, causing paradoxes - so it's very unlikely that this is possible.

• so the time it takes for light to reach the person is dependent on the location of the object not the speed?
– John
Feb 26, 2012 at 17:39
• Yes, the time taken for the light to reach you is simply (distance to object)/(speed of light), regardless of the speed at which you or the object are moving. Feb 26, 2012 at 18:26

The issue of tachyonness is a red herring; and the question is entirely a matter of simple computation, that you should do for yourself.

Here it is.

Let $$x = v t$$ describe the motion of an object in a direction parallel to the $$x$$ axis, where $$t$$ is the time, where the origin on the $$x$$ axis is where you are at and where time $$t = 0$$ is when the object is where you are at. It is moving with a speed $$v$$, which is negative if it is toward you, positive if it is moving away from you.

We'll also assume you're staring in the direction of increasing $$x$$. A signal moving at the light speed $$c$$ has a motion given by $$x = b - c t$$, if it is coming at you from the direction of increasing $$x$$. If it is at the location of the object at $$(x, t) = \left(x_0, t_0\right)$$ and at your location at $$(x, t) = \left(x_1, t_1\right)$$, then $$x_0 = b - c t_0, \hspace 1em x_0 = v t_0,\\ x_1 = b - c t_1, \hspace 1em x_1 = 0.$$ Therefore $$(v + c) t_0 = b = c t_1$$, and: $$(v + c) x_0 = (v + c) v t_0 = v (v + c) t_0 = v c t_1.$$

The motion you see is given by the stream of signals and is therefore described by $$(x,t) = \left(x_0, t_1\right)$$. You could do this extra carefully by trying to separately account for left and right eyes and use triangulation to determine the "apparent" distance in place of $$x_0$$, but I don't think that extra complication will significantly affect the issue (maybe except when it's close), so we'll just use $$x_0$$, itself.

Thus, assuming $$v ≠ -c$$, and $$v ≠ 0$$, we have $$x_0 = \frac{v c}{v + c} t_1 = \frac{c}{1 + c/v} t_1 = \frac{v}{1 + v/c} t_1,$$ and you see the object moving with a speed $$c/(1 + c/v) = v' = v/(1 + v/c)$$.

If $$v = 0$$, then $$x_0 = v t_0 = 0$$ and you see the object up right next to you, so $$v' = 0$$.

If $$v = -c$$, then $$b = 0$$ and, thus, $$t_1 = 0$$, and that's the only signal you'll ever see, and you'll only see it in a flash at time 0, and $$v' = ∞$$.

If $$v > 0$$, then $$0 < v' < c$$. If $$-c < v < 0$$ then $$v' < 0$$. If $$v < -c$$ then $$v' > 0$$. If $$v = ∞$$ then $$v' = c$$. An object going at the speed $$v = ∞$$ will appear to you to be moving away from you at light speed - even if you turn around and look at it in the other direction.

By the way, notice that I actually said nothing about Relativity in the determination of $$v'$$. Relativity is also a red herring to the question; it's the same answer as in non-relativistic physics. You get the same expression for $$v'$$, regardless of which paradigm you are in, it is paradigm-independent, and the question actually has nothing to do with Relativity, per se, but merely with signal processing. You could just as well take any other speed $$V$$ in place of $$c$$ (e.g. the speed of light in water) to determine the visual effect of an object's motion, in which case you'll get $$v/(1 + v/V) = v' = V/(1 + V/v)$$ in place the expression previously derived - again, independent of paradigm. So, even light speed is a red herring here.