See subject. I still can’t get my head around sometimes mass/sometimes no-mass entities to begin with. But I have heard this point about photons made, so it occured to me that we must be bearing a heavier load under more light.
Thanks,
Leo
They don’t have mass, but they do have momentum, and so yes, light exerts a pressure on surfaces upon which it is incident. The sun is pushing you away from it (e.g. toward the ground when the sun is high in the sky).
The question then becomes whether you can feel the pressure. The answer is no. It’s many orders of magnitude too small. In fact, I doubt we have any real world equipment that could measure the difference on an actual human being. If I’m wrong about that, please link. I’d love to see it.
Well, the radiation pressure of direct sunlight is 0.000005 N/m^2 on earth. That is about 0.0000011 pounds/m^2.
Guess it depends on how much of a profile you present to the sun (standing, lying down, etc.) to calculate how much more you weigh.
The smallest weight ever measured is a “few billionths of a trillionth of a gram - or a few zeptograms”. So I suppose we could measure this.
I didn’t ask if there were a laboratory scale that could measure such a weight. I asked if the difference could be measured on a living human being.
Well, a square meter on earth feels a pressure of 0.000498 grams. That is pretty small but measurable. I will assume a human laying down presents at least a square meter of surface area (just a gut feeling…I do not know how to calculate that).
Does a scale exist that could weigh a human and note that difference? I doubt it. I cannot see why a scale that could measure the weight of a human (say 80 kilos) would need such precision.
Is it theoretically possible to build a scale that could weigh a human to such precision? Again I have no idea but I suspect it could be done if someone really wanted to.
The experiment would be easy enough. Place a human in the sun on a clear day with the sun directly overhead and measure his weight. Put a cardboard box over him to block all light and measure his weight again.
Depends how bright you are?
Grams are not a unit of pressure.
A photon is always massless. But a system of photons generally isn’t – the reason for that is that mass of an object is determined by the equation E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup], the more familiar form of which, E = mc[sup]2[/sup], is of course obtained when p = 0, i.e. when the object has no momentum, or equivalently is at rest.
Now, this gives for the mass: m[sup]2[/sup]c[sup]4[/sup] = E[sup]2[/sup] - p[sup]2[/sup]c[sup]2[/sup], which for an individual photon is equal to zero. But for, say, two photons, we have m[sup]2[/sup]c[sup]4[/sup] = (E[sub]1[/sub] + E[sub]2[/sub])[sup]2[/sup] - (p[sub]1[/sub] + p[sub]2[/sub])[sup]2[/sup]c[sup]2[/sup], which no longer has to be (you can check this yourself just plugging in numbers, with total disregard for their meaning – say (E[sub]1[/sub], p[sub]1[/sub]) = (5, 5/c), and (E[sub]2[/sub], p[sub]2[/sub]) = (3, -3/c); clearly, in both cases, E[sup]2[/sup] - p[sup]2[/sup]c[sup]2[/sup] = 0, but (E[sub]1[/sub] + E[sub]2[/sub])[sup]2[/sup] - (p[sub]1[/sub] + p[sub]2[/sub])[sup]2[/sup]c[sup]2[/sup] = (5 + 3)[sup]2[/sup] - (5 - 3)[sup]2[/sup] = 60*).
This ‘invariant mass’ is more than just an abstract quantity – it is the actual mass the system can ‘use’ to produce massive particles – or equivalently, if a massive particle decays into massless ones, its (rest) mass will be equal to the invariant mass of the resulting system of massless particles.
So you won’t have more mass due to the photons, but the system of you + photons will have an invariant mass greater than your rest mass.
*You may have noticed that if I had used all positive quantities, we’d have gotten 0 again, so perhaps you’re wondering if I didn’t pull a fast one on you by including a negative quantity. But the reason is that actually, p is a vector quantity, i.e. something with both direction and magnitude, and only if the directions of both momenta align – if the particles move in parallel – is the invariant mass zero. But since I was using numbers only, the only possibility to have the right magnitude, but different direction, was to use the negative quantity – corresponding to both particles moving antiparallel to one another.
Cecil talked about the inverse problem (more or less) a few months back.
it’s not a photon momentum question. it’s a gravity question; whether the exposed sun exerts greater gravity than one hiding behind the ground (nightime.)
The problem with this method is inability to isolate your system. A few dust mites, or losing a bead of sweat, or a wandering mosquito could potentially screw with precision that small.