If you were in a car doing the speed of light and you turned the lights on, would they do anything?

Smart-aleck answer: Yes. Somewhere above Mach 1 (a very tiny fraction of the speed of light,) the glass in the lights would melt as the car falls apart and burns up while accelerating.

Answer you probably want: No. If you had a magic car that traveled at "c" - when you turn on the headlights at the speed of light (assuming you mean the headlights and not the inside dome light, but even then...) the headlight beam would go at the speed of light relative to the magic car.

Since this is the "bad movie" page: In the movie "The Last Starfighter," actor Robert Preston drives just such a car. (Its a disguised alien spaceship.)

<font size=-1>[ This Message was edited by: Chip on 2002-08-06 17:53 ]</font>

So to a stationary observer the headlight beams would be traveling at twice the speed of light. Is that right? Wh have discussed thes before with my nose (long story) but i don't rmemeber with headlights.

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GIYUL :-)
"It takes Thousands to fight a battle for a mile, Millions to hold an election for a nation, but it only takes One to change the world." - Dan Sandler 2002

Well, actually, Relativity says the car can't travel at the speed of light, so if you're talking about a car actually going that speed, you're talking about a non-relativistic universe, which means anything could happen.

If you are in a car travelling slightly slower than c (which is allowed in relativity), and you turn on the headlights, then the driver of the car would see the headlight beams as travelling away from the car at the speed of light. The stationary observer would see the beams travelling at the speed of light relative to himself, and so would say that they were not moving that fast relative to the car.

Simple, ain't it? [img]/phpBB/images/smiles/icon_biggrin.gif[/img]

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SeanF

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The contents of this post are ©2008 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

Nope... To the stationary observer, the car, travelling at the speed of light, would emit a photon of light, also travelling at the speed of light. The two would appear to travel together.

Suppose the car turned on its turn-signals: i.e., emitting light at 90 degrees to its path of travel: the light would travel at c, not at c times the square root of two, as one might expect from vector addition.

Suppose the car flashes its brake-lights: i.e., emitting light backwards. The light would travel at c, not at a speed of zero, as one might expect from vector addition.

Given these ideas, the Lorentz-Fitzgerald equations fall out.

Silas

Thanks Silas and SeanF. I am actually getting it. That really makes sense!! [img]/phpBB/images/smiles/icon_smile.gif[/img] YAY!! I am getting the basics of Physics, now onto thermodynamics, Quantum physics, and the hardest of all Algebra!!:-)

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<font size=-1>[ This Message was edited by: g99 on 2002-08-07 00:39 ]</font>

<font size=-1>[ This Message was edited by: g99 on 2002-08-07 00:40 ]</font>

One point you seem to be forgetting. If you were somehow able to travel at the speed of light, you would be frozen in motion from the perspective of other observers. So if you are as stiff as a statue, then how could you turn the lights on ?

Phobos

<font size=-1>[ This Message was edited by: Phobos on 2002-08-07 00:57 ]</font>

____________Remote control [img]/phpBB/images/smiles/icon_smile.gif[/img]_____
"The chickens is coming!!!"
"Watch out for falling coconuts!!"
The creationist dogma: "If you can't prove it I must be right"

<font size=-1>[ This Message was edited by: g99 on 2002-08-07 14:51 ]</font>

Another thing people keep forgetting is the doppler effect. Since light can't go faster than c, the energy gained from momentum instead ****s the wavelength up towards the bluer end. So the light from your headlights is still at c, but is shifted way up into the gamma-ray end of the spectrum.

Conversely, light from your taillights lose energy, and therefore get downshifted, they'd fall down into the radio-wave end of the spectrum.

There are actually several differest effects experienced as you approch c. This page is pretty good at explaining them:

http://www.fourmilab.ch/cship/cship.html

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The operative phrase here is "from the perspective of other observers". Time only seems to stand still for those outside of the car. You, however are not affected by that, so you just reach up and flick on the lights as normal. It just an infinite amount of time to do it.

Actually, this is all only true for speeds approaching c. As mentioned before, nothing with mass can reach c, and all effects mentioned stretch either to infinity or to zero. It's impossible to say anything would happen because the theory doesn't allow for it. It's an impossible situation.

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...And that, my liege, is how we know the Earth to be banana-shaped. --Sir Bedevere

Question from the "relativity clueless":

Would someone please explain again why something with mass cannot reach c?

Ignouring the Relativity bad Pun: [img]/phpBB/images/smiles/icon_rolleyes.gif[/img]

Simple Answer:
Think of it like Pushing a Rock up a Mountain (Dang that Red Bull comercial), except that This Mountain is Special!

Although it Starts Off shallow, The Curve Steepens towards The End, with the Grade becoming 2/1 when you've gone 86.60% of The Way, 3/1 when you've reached 94.28%, [/b]7/1[/b] when you've hit 98.97%, and So On, and So Forth, with it Coming Closer to The Vertical, The Closer that you Get to The End, with it Becoming Infinitely Steep, not to be Confused with Vertical, by The End.

Big Nasty, Grab your Ears before your Brain Leaks out, Answer:

It's a Variation of The Pythagorean Theorem, known as The Lorentz-Fitzgerald Equation, after its Discoverers, it is Derived as Follows:

a^2=c^2-b^2

a=x=The Distance Traveled | b=v=The Velocity | c=The Speed of Light:

x^2=c^2-v^2

Divide both Sides by c^2:

x^2/c^2=1-v^2/c^2

Square Root both Sides:

x/c=SqRt(1-v^2/c^2)

This Version is used to Figure out Length Contraction in The Following Form | l'=The length Observed by the Moving Observer | l=The length Measured When at Rest with Respect to the Object:

l'=l*SqRt(1-v^2/c^2)

For Time, and Mass, Calculations, it Must be Inverted:

c/x=1/SqRt(1-v^2/c^2)

Often expressed in the Following Forms:

t'=The length Observed by the Moving Observer | t=The length Measured When at Rest with Respect to the Object:

t'=t/SqRt(1-v^2/c^2)

m'=The length Observed by the Moving Observer | m=The length Measured When at Rest with Respect to the Object:

m'=m/SqRt(1-v^2/c^2)

If we Set m Equal to 1, you Get The much Simpler:

m'=1/SqRt(1-v^2/c^2)

Just Remember though, you can Enter The Velocity as a Percentage, as it Saves you from Having to Type 299,792.458 km, All the Time!

Above All, HAVE FUN! [img]/phpBB/images/smiles/icon_biggrin.gif[/img]

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<font size=-1>[ This Message was edited by: ZaphodBeeblebrox on 2002-08-09 04:37 ]</font>

What Zaphod seems to be saying in so many words is that, the closer you get to c, the more energy it takes to push a mass faster. And at c itself, the energy required for greater acceleration becomes infinite.

Therefore nothing with mass can reach or exceed the speed of light. Only something massless like photons can do that.

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...And that, my liege, is how we know the Earth to be banana-shaped. --Sir Bedevere

My simple thoughts on the energy required to accelerate to the speed of light. I think of it as "cosmic wind resistance". Similar to wind resistance here on Earth. The faster your speed the more "cosmic wind resistance" you encounter hence more energy is required to accelerate you per unit of speed than the last unit. So at from .05c to .1c not much resistance but from .9c to .95c good luck. Just my thoughts.

P.S. Just out of curiosity taking the energy required to accelerate 1Kg from .9c to .95c how much could you accelerate (in terms of c) another (different) 1Kg from rest using the same amount of energy? Please assume objects are in space and not encountering any gravitational or other resistance.

-AJ

Careful with that AJ, that was similar to the arguments used to prove that we couldn't break the sound barrier, I think.

Good point Grapes although the sound barrier was broken quite a while before my time (October 14, 1947). Interestingly enough I still remember my Calculus teacher almost cursing the engineers of that age for not following the airflow equations out to large enough values where they would have noticed a discontinuity in the function. He was quite upset at the number of test pilots who died because of this mathematical oversight. It certainly made me double check my Calculus answers.