Length contraction and Time dilation for LIGHT?

[Stephen Bulking]

Homework Statement

A message is sent via radio wave (λ = 10 m) from Earth to a nearby star, Proxima Centauri, which lies 1.3 parsecs from Earth. A parsec is a unit of length and 1 parsec = 3.1 × 10^16m. How long will it take the message to reach Proxima Centauri traveling through vacuum?
(a) 0 yrs, the message arrives instantly.
(b) 1.38 yrs
(c) 4.26 yrs
(d) 255.0 yrs
(e) 1.38 × 108 yrs

Relevant Equations

t = t0/γ (time dilation)
l = γl0 (length contraction)
t0: proper time
l0: proper length

Radio wave travels at the speed of light 3x10^8 (m/s)
Converting the distance to meter: 1.3 x 3.1x 10^16 = 4.03x10^16m
The time it takes in our Earth frame of reference is: 4.03x10^16m/3x10^8 (m/s) = 4.26 years
The answer is B
But wouldn't the time in light's frame of reference be 0 and it's length be infinite?
If it was another particle traveling at near light speed than it it's frame of reference, the time and distance it has to travel are still given by time dilation and length contraction formulas right?

 

[etotheipi]

The question is just asking for $\Delta x = c \Delta t$, nothing more.

There's no such thing as the frame of reference of light! But for a particle with $v$ very close to $c$, everything works just fine.

 

[PeroK]

But wouldn't the time in light's frame of reference be 0 and it's length be infinite?
If it was another particle traveling at near light speed than it it's frame of reference, the time and distance it has to travel are still given by time dilation and length contraction formulas right?

The question ought to state that these measurements are in the rest frame of the Earth.

There is no frame of reference for light, or inertial reference frame moving at $c$ with respect to another.

The proper time for a particle (i.e. in its reference frame) moving at near the speed of light in the Earth's reference frame would indeed by less - and, in fact, can be made arbitraily small; but not zero.

 

[jbriggs444]

traveling through vacuum?

If you needed one, there is a hint there. The message is "traveling". If the message were stationary while the endpoints moved then the endpoints would have been "traveling".

It is sometimes useful to look at the problem statement to see if a particular frame of reference is implied.

 

[Vanadium 50]

But wouldn't the time in light's frame of reference be 0 and it's length be infinite?

Whether or not that is well-defined (and it's not), at no point were you asked that.

 

[vela]

Radio wave travels at the speed of light 3x10^8 (m/s)
Converting the distance to meter: 1.3 x 3.1x 10^16 = 4.03x10^16m
The time it takes in our Earth frame of reference is: 4.03x10^16m/3x10^8 (m/s) = 4.26 years
The answer is B.

B? Not C?

 

[Stephen Bulking]

B? Not C?

Yeah man I got B, there's no given answer to check with but from my calculations I got B

 

[PeroK]

Yeah man I got B, there's no given answer to check with but from my calculations I got B

You said you got 4.26 years, which is option C.

 

[Stephen Bulking]

You said you got 4.26 years, which is option C.

Oh right, sorry, typo.