# Why do light, radio and electricity all travel at the same speed?

**URL:** <https://boards.straightdope.com/t/why-do-light-radio-and-electricity-all-travel-at-the-same-speed/299403>\
**Category:** Factual Questions\
**Created:** [April 15, 2005, 1:29am UTC](https://boards.straightdope.com/t/why-do-light-radio-and-electricity-all-travel-at-the-same-speed/299403 "2005-04-15T01:29:46Z")\
**Posts on this page:** 5\
**Page:** 3

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**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\
**Post date:** [April 17, 2005, 8:19am UTC](https://boards.straightdope.com/t/why-do-light-radio-and-electricity-all-travel-at-the-same-speed/299403/41 "2005-04-17T08:19:06Z")

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> [@Ring](#):
>
> That certainly doesn’t surprise me.

Well, it surprises _me_. OK, here’s an equation I’m familiar with:

lambda = 2.pi.delta, where

lambda = wavelength [m],  
pi = 3.14159… [dimensionless, constant],  
delta = skin depth = sqrt(1/(pi.f.u.sigma)) [m]  
f = frequency [Hz],  
u = permeability [Fm[sup]-1[/sup]],  
sigma = conductivity [Sm[sup]-1[/sup]].

Now, for your example of a 1 MHz wave in copper:

f = 10[sup]6[/sup] Hz  
sigma = 5.7x10[sup]7[/sup] Sm[sup]-1[/sup]  
u[sub]r[/sub] = 1

lambda = 2 x 3.14159 x sqrt(1/(3.14159 x 10[sup]6[/sup] x 4 x 3.14159 x 10[sup]-7[/sup] x 5.7 x 10[sup]7[/sup])) = 4.2x10[sup]-4[/sup] m

Now,

v = lambda.f, where

v = propagation speed [ms[sup]-1[/sup]]

v = 4.2x10[sup]-4[/sup] x 10[sup]6[/sup] = 420 ms[sup]-1[/sup]

So, OK, I believe you. David Pace’s figure of 408 ms[sup]-1[/sup] can be achieved by using sigma = 6x10[sup]7[/sup] Sm[sup]-1[/sup] (as David Pace did).

This is really weird. David Pace states “Also, the permittivity may be approximated as e[sub]0[/sub]”. That’s what I’d always though, but it can’t possibly be right.

The propagation velocity of 420 ms[sup]-1[/sup] implies a different value for e[sub]r[/sub].

420 = 1/sqrt(u e) = 1/sqrt(u[sub]0[/sub]u[sub]r[/sub] e[sub]0[/sub]e[sub]r[/sub])  
= 1/sqrt(4 x 3.14159 x 10[sup]-7[/sup] x 8.854x10[sup]-12[/sup] x e[sub]r[/sub])  
e[sub]r[/sub] = 5.1x10[sup]11[/sup]

This confirms **CurtC** ’s earlier statement:

> [@CurtC](#):
>
> I’m pretty sure that the e[sub]r[/sub] for copper, or any conductor, is extremely high. So high that it’s assumed to be infinite in most calculations. I’m trying to find a reference…

(No need for that reference, now, Curt. 🙂 )

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**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\
**Post date:** [April 17, 2005, 8:38am UTC](https://boards.straightdope.com/t/why-do-light-radio-and-electricity-all-travel-at-the-same-speed/299403/42 "2005-04-17T08:38:57Z")

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> [@Ring](#):
>
> Another thought: The electric field vector inside the conductor would point in the direction of current flow which means the Poynting vector would be perpendicular to the surface of the conductor. So this _“inside the conductor wave”_ must be what causes the Ohmic heating?

I think this is getting unnecessarily complicated. Ohmic heating occurs at zero frequency, so there’s some fundamental cause of ohmic heating unrelated to wave propagation, and unrelated to Maxwell. A kinetic model makes more sense, i.e. moving electrons transferring energy to atoms in the conductor.

At non-zero frequencies, Maxwell can tell you how the current will be distributed within the conductor, i.e. current density is higher near the surface, and decreases exponentially with distance from the surface. But this can’t be described as the “cause” of the Ohmic heating.

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**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\
**Post date:** [April 18, 2005, 6:40am UTC](https://boards.straightdope.com/t/why-do-light-radio-and-electricity-all-travel-at-the-same-speed/299403/43 "2005-04-18T06:40:02Z")

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I’ll yield on the propagation speed through (or rather, around) wires… It’s been a few years since I’ve taken E&M, and it would take a lot greater provocation than a message board for me to pull my Jackson back off the shelf. I do, however, recall that one can derive the Ohmic heating from Poynting flux, even for DC (this showed up as a homework problem in both undergrad and graduate E&M).

As to the speed of gravity, the biggest reason to say that it’s c is that that’s the only speed in the Universe. If you have two different speeds inherent to physics, then Lorentz invarience goes out the window, and takes all of relativity with it. The speed of light is the same in all reference frames, but for any other speed, you have to specify a reference frame. So if you said, for instance, that the speed of gravity is .1 c, or 100 c, or instantaneous, or zero, you would have to say what reference frame that was in.

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**Author:** ![Askance](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/askance/32/8281_2.png) [@Askance](https://boards.straightdope.com/u/Askance)\
**Post date:** [April 18, 2005, 6:51am UTC](https://boards.straightdope.com/t/why-do-light-radio-and-electricity-all-travel-at-the-same-speed/299403/44 "2005-04-18T06:51:08Z")

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> [@Roches](#):
>
> Eleusis: Elephant rhinoceros sin _θ_ in a direction mutually perpendicular to the elephant and the rhinoceros as determined by the right hand rule.
> 
> What do you get when you cross an elephant with a mountain climber?
> 
> You can’t; a mountain climber is a scalar.

What’s a polar bear?

It is a usual, rectangular bear, transformed to the polar coordinates

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**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\
**Post date:** [April 18, 2005, 3:22pm UTC](https://boards.straightdope.com/t/why-do-light-radio-and-electricity-all-travel-at-the-same-speed/299403/45 "2005-04-18T15:22:41Z")

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> [@Chronos](#):
>
> it would take a lot greater provocation than a message board for me to pull my Jackson back off the shelf.

As long as you leave your Johnson up there.

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